A2 October 2021 Q4

EdexcelCurrent spec11 marksRecurrence Relations

4. Sequences \(\{x_n\}\) and \(\{y_n\}\) for \(n \in \mathbb{N}\), are defined by

\[x_{n+1} = 2y_n + 3 \quad \text{and} \quad y_{n+1} = 3x_{n+1} - 4x_n\]\[x_1 = 1 \quad \text{and} \quad y_1 = a\]

where \(a\) is a constant.

(a) Show that \(x_{n+2} - 6x_{n+1} + 8x_n = 3\) (1)

(b) Solve the second-order recurrence relation given in (a) to obtain an expression for \(x_n\) in terms of \(a\) and \(n\). (8)

Given that \(x_7 = 28\,225\)

Mark scheme (a)

Scheme	Marks	AO
\(x_{n+2} = 2y_{n+1} + 3 \Rightarrow x_{n+2} = 2(-4x_n + 3x_{n+1}) + 3\) Leading to \(x_{n+2} - 6x_{n+1} + 8x_n = 3\,*\)	B1	2.2a
	(1)

B1: Correct reasoning to derive given result – sufficient working must be shown as recurrence equation given in question

Mark scheme (b)

Scheme	Marks	AO
aux equation \(m^2 - 6m + 8 = 0 \Rightarrow m = 2, m = 4\)	B1	2.1
\(x_n = A(2)^n + B(4)^n\)	B1	1.1b
particular solution try \(x_n = \lambda\) \(\therefore \lambda - 6\lambda + 8\lambda = 3 \Rightarrow \lambda\,[= 1]\)	M1	1.1b
\(x_n = A(2)^n + B(4)^n + 1\)	A1	2.2a
\(x_1 = 1 \Rightarrow 2A + 4B = 0\)	M1	1.1b
\(y_1 = a \Rightarrow x_2 = 2a + 3\)	B1	3.1a
\(4A + 16B + 1 = 2a + 3\)	M1	1.1b
\(A = -\dfrac{(a+1)}{2},\ B = \dfrac{(a+1)}{4} \Rightarrow x_n = (a+1)(4)^{n-1} - (a+1)(2)^{n-1} + 1\) (oe)	A1	2.2a
	(8)

B1: cao for auxiliary equation and corresponding solutions (this mark can be implied by the correct complementary function)

B1: cao for the complementary function

M1: Substitute \(x_n = \lambda\) into their second order recurrence relation and solve for \(\lambda\)

A1: Correct general solution

M1: Forms one equation in \(A\) and \(B\) using \(x_1 = 1\)

B1: Uses original recurrence relation for \(x_{n+1}\) to derive the expression \(2a + 3\)

M1: Setting up a second equation in \(A\) and \(B\)

A1: cao (oe e.g., \(x_n = 0.25(a+1)(4)^n - 0.5(a+1)(2)^n + 1\))

Mark scheme (c)

Scheme	Marks	AO
As \(x_7 = 28225 \Rightarrow (a+1)(4)^6 - (a+1)(2)^6 + 1 = 28225\) leading to \(a = \ldots\)	M1	3.4
\(a = 6\)	A1	2.2a
	(2)
	(11 marks)

M1: Using \(x_7 = 28225\) to form a linear equation in \(a\) and attempt to solve for \(a\)

A1: cao for \(a\)